Prove that for a class of sets we have,
Proof. Let be an arbitrary element of . This means that and , which means that is not in for any in the class . Hence, for every we have (since is in no matter what, and is not in for any that we choose, so it must be in ). But, for all means that is in the intersection ; and hence, .
For the reverse inclusion, let be any element in . This means that for every , i.e., and for every . Since for every , we then have . Hence, . Therefore, .
Hence, . ∎
Proof. Let be any element of . This means that and . Further, not in the intersection of the sets means that there is at least one , say , such that . Since and , we know . Then we can conclude . Thus, .
For the reverse inclusion, we let be any element in . This means there is at least one , say , such that , which means and . Since , we know ; therefore, . Hence, .